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Finite population correction factor

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Whenever we take a random sample from a population having variance σ, we know that the sample mean has variance given as σ/√n where n is the sample size. But this is only true if we assume that:

  1.  The sample has been chosen with replacement.
  2.  The population size is infinite.

Both of these assumptions are false in real-life situations. But the above estimate for the variance of the sampling distribution still works as an approximation as long as the sample size is less than 5% of the total population.

If the sample size exceeds 5% of the population then we need to multiply the standard deviation of the sample mean by a correction factor called the FPC (Finite population correction factor).

Formula for the finite population correction factor:

The FPC can be calculated using the formula,

Finite Population Correction Factor formula

Thus the formula for standard error of sampling mean becomes,

Corrected formula for variance of sample mean using Finite Population Correction Factor

And the formula for standard error of sampling proportion becomes,

Corrected formula for variance of sample proportion using Finite Population Correction Factor

Example 1: Suppose that a population of size 10000 has a variance of 100. We want to estimate the average height of people in the population. If a sample of size 2500 is selected from the population then calculate the standard error of the sample mean.

Solution: The formula for standard error is given as S.E = σ/√n but since the sample is of size greater than 5% of the population (=500) therefore we must apply the correction factor to the standard error.

The formula becomes, S.E = √[(N-n)/(N-1)]* σ /√n = √(7500/9999) * 10/50 = 0.17321

Example 2: Suppose that we have a population of size 1000 that has a variance of 25. We obtain an estimate for the proportion ‘p’ of blue-eyed people in the population as p=0.5. If we have drawn a sample of size 100 from the population then calculate the standard error of the population proportion

Solution: The formula for standard error is given as, S.E = √p(1-p)/n but since the sample is of size greater than 5% of the population (=50) therefore we must apply the correction factor to the standard error.

The formula becomes, S.E = √[(N-n)/(N-1)]* √p(1-p)/n = √(900/999) * √(0.5*0.5/100) = 0.0474

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